Question

Determine the specific volume of superheated water vapor at \(15 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\), using \((a)\) the ideal-gas equation, ( \(b\) ) the generalized compressibility chart, and (c) the steam tables. Also determine the error involved in the first two cases.

Short answer

Answer: The specific volume using the ideal-gas equation is 0.01915 m³/kg, using the generalized compressibility chart is 0.01723 m³/kg, and using steam tables is 0.01717 m³/kg. The error involved in the ideal-gas equation is 11.52%, and the error in the generalized compressibility chart method is 0.35%.

Step-by-step solution

Specific volume using the ideal-gas equation

To determine the specific volume using the ideal-gas equation, we use the following equation: \(v = \frac{RT}{P}\) where \(v\) is the specific volume, \(R\) is the specific gas constant for water vapor, \(T\) is the temperature in Kelvin, and \(P\) is the pressure. First, we need to convert the given temperature and pressure to appropriate units: Temperature: \(T = 350^{\circ}C = 350 + 273.15 = 623.15 K\) Pressure: \(P = 15 \, \text{MPa} = 15 \times 10^6 \, \text{Pa}\) R = 461.5 J/(kg*K) (Specific gas constant for water vapor) Now, we can calculate the specific volume using the ideal-gas equation: \(v = \frac{RT}{P} = \frac{461.5 \, \text{J/(kg*K)} \times 623.15 \, \text{K}}{15 \times 10^6 \, \text{Pa}} = 0.01915\,\text{m}^3/\text{kg}\)

Specific volume using the generalized compressibility chart

To determine the specific volume using the generalized compressibility chart, we need to find the compressibility factor, \(Z\), at the given reduced pressure and reduced temperature. The reduced pressure and temperature can be calculated using: \(P_r = \frac{P}{P_c}\) \(T_r = \frac{T}{T_c}\) For water vapor, the critical pressure \(P_c = 22.064 \, \text{MPa}\) and the critical temperature \(T_c = 647.3 \, \text{K}\). We can now determine the reduced pressure and temperature: \(P_r = \frac{15 \, \text{MPa}}{22.064 \, \text{MPa}} = 0.68\) \(T_r = \frac{623.15 \, \text{K}}{647.3 \, \text{K}} = 0.963\) Using the generalized compressibility chart, we find the compressibility factor \(Z\) at these values to be approximately \(Z=0.9\). Now, we can find the specific volume using the following equation: \(v = Z \frac{RT}{P} = 0.9 \times \frac{461.5 \, \text{J/(kg*K)} \times 623.15 \, \text{K}}{15 \times 10^6 \, \text{Pa}} = 0.01723 \, \text{m}^3/\text{kg}\)

Specific volume using steam tables

Using steam tables, we look up the specific volume at the given pressure and temperature. For water vapor at \(15 \, \text{MPa}\) and \(350^{\circ} C\), we find the specific volume to be: \(v = 0.01717 \, \text{m}^3/\text{kg}\)

Error calculation

Now we can determine the error involved in the ideal-gas equation and the generalized compressibility chart: Error in ideal-gas equation: \(error = \frac{|v_{\text{ideal}}-v_{steam}|}{v_{steam}} \times 100 = \frac{|0.01915-0.01717|}{0.01717} \times 100 = 11.52 \%\) Error in generalized compressibility chart: \(error = \frac{|v_{\text{chart}}-v_{steam}|}{v_{steam}} \times 100 = \frac{|0.01723-0.01717|}{0.01717} \times 100 = 0.35 \%\) The error in the ideal-gas equation method is 11.52%, and the error in the generalized compressibility chart method is 0.35%.